BdMO National Higher Secondary 2019/3
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Let $\alpha$ and $\omega$ be two circles such that $\omega$ goes through the center of $\alpha$.$\omega$ intersects $\alpha$ at $A$ and $B$.Let $P$ any point on the circumference $\omega$.The lines $PA$ and $PB$ intersects $\alpha$ again at $E$ and $F$ respectively.Prove that $AB=EF$.
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Re: BdMO National Higher Secondary 2019/3
Let $O$ be the center of $w_1$ and define $OP \cap AB$ as $Q$
Consider a circular inversion with respect to $w_1$
It is obvious that $AB$ is the inverse of $w_2$ and $Q$ is the inverse of $P$ with respect to $w_1$
$\therefore \angle APO = \angle BPO = \angle ABO= \angle BAO$ and $\angle ABF= \angle AEF$
Since, $APOF$ is cyclic so $\angle OBF = \angle EAO = \angle OEA = \angle OFB$
Also,$\angle BOF = \angle EOA$ so $\angle APF + angle AOB = \angle APF + \angle EOF = 180°$
$\Longrightarrow$ $EOPF$ is cyclic and $Q$ is the radical center of $w_1, w_2, \odot (EOPF)$
$\Longrightarrow E, P, F$ are collinear
MOREOVER, $POWw_1(P)=PE.PA = PB.PF \Longrightarrow QA.QE = QB.QF$( inversion formula and POP)
Also $POWw_1(Q) =QE.QF = QA.QB$
Combining these results yields $AQ = FQ $ and $ BQ = EQ$
$\Longrightarrow AQ + BQ = EQ + FQ$
$\Longrightarrow AB = EF$ (A.W.D)
Consider a circular inversion with respect to $w_1$
It is obvious that $AB$ is the inverse of $w_2$ and $Q$ is the inverse of $P$ with respect to $w_1$
$\therefore \angle APO = \angle BPO = \angle ABO= \angle BAO$ and $\angle ABF= \angle AEF$
Since, $APOF$ is cyclic so $\angle OBF = \angle EAO = \angle OEA = \angle OFB$
Also,$\angle BOF = \angle EOA$ so $\angle APF + angle AOB = \angle APF + \angle EOF = 180°$
$\Longrightarrow$ $EOPF$ is cyclic and $Q$ is the radical center of $w_1, w_2, \odot (EOPF)$
$\Longrightarrow E, P, F$ are collinear
MOREOVER, $POWw_1(P)=PE.PA = PB.PF \Longrightarrow QA.QE = QB.QF$( inversion formula and POP)
Also $POWw_1(Q) =QE.QF = QA.QB$
Combining these results yields $AQ = FQ $ and $ BQ = EQ$
$\Longrightarrow AQ + BQ = EQ + FQ$
$\Longrightarrow AB = EF$ (A.W.D)